Question
Draw a $\triangle\text{ABC}$ in which base BC = 6cm, AB = 5cm and $\angle\text{ABC}=60^\circ.$ Then construct another triangle whose sides are of the corresponding sides of $\triangle\text{ABC}.$
✓
Answer
Steps of construction:
i. Draw a triangle $A B C$ with side $B C=6 cm, A B=5 cm$ and $\angle A B C=60^{\circ}$.
ii. Draw a ray BX , which makes an acute angle $\angle CBX$ below the line BC .
iii. Locate four points $B_1, B_2, B_3$ and $B_4$ on $B X$ such that $B_1=B_1 B_2=B_2 B_3=B_3 B_4$.
iv. Join $B_4 C$ and draw a line through $B_3$ parallel to $B_4 C$ intersecting $B C$ to $C^{\prime}$.
v. Draw a line through $C^{\prime}$ parallel to the line $C A$ to intersect $B A$ at $A^{\prime}$.
Justification of the construction:$\because\text{B}_4\text{C }||\text{ B}_3\text{C}'$ [By construction]
$\therefore\ \frac{\text{BB}_3}{\text{BB}_4}=\frac{\text{BC}'}{\text{BC}}$
[Basic Proportionality Theorem] But $\frac{\text{BB}_3}{\text{BB}_4}=\frac{3}{4}$ [By condtruction]$\therefore\ \frac{\text{BC}'}{\text{BC}}=\frac{3}{4}\ \dots(\text{i})$
$\because\text{CA }||\text{ C}'\text{A}'$ [By construction]
$\therefore\ \triangle\text{BC}'\text{A}'\sim\triangle\text{BCA}$
From equation (i),$\frac{\text{A}'\text{B}}{\text{AB}}=\frac{\text{A}'\text{C}'}{\text{AC}}=\frac{\text{BC}'}{\text{BC}}=\frac{3}{4}$
[Basic Propotionality Theorem]
i. Draw a triangle $A B C$ with side $B C=6 cm, A B=5 cm$ and $\angle A B C=60^{\circ}$.
ii. Draw a ray BX , which makes an acute angle $\angle CBX$ below the line BC .
iii. Locate four points $B_1, B_2, B_3$ and $B_4$ on $B X$ such that $B_1=B_1 B_2=B_2 B_3=B_3 B_4$.
iv. Join $B_4 C$ and draw a line through $B_3$ parallel to $B_4 C$ intersecting $B C$ to $C^{\prime}$.
v. Draw a line through $C^{\prime}$ parallel to the line $C A$ to intersect $B A$ at $A^{\prime}$.
Justification of the construction:$\because\text{B}_4\text{C }||\text{ B}_3\text{C}'$ [By construction]
$\therefore\ \frac{\text{BB}_3}{\text{BB}_4}=\frac{\text{BC}'}{\text{BC}}$
[Basic Proportionality Theorem] But $\frac{\text{BB}_3}{\text{BB}_4}=\frac{3}{4}$ [By condtruction]$\therefore\ \frac{\text{BC}'}{\text{BC}}=\frac{3}{4}\ \dots(\text{i})$
$\because\text{CA }||\text{ C}'\text{A}'$ [By construction]
$\therefore\ \triangle\text{BC}'\text{A}'\sim\triangle\text{BCA}$
From equation (i),$\frac{\text{A}'\text{B}}{\text{AB}}=\frac{\text{A}'\text{C}'}{\text{AC}}=\frac{\text{BC}'}{\text{BC}}=\frac{3}{4}$
[Basic Propotionality Theorem]
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